The number of turns in primary and secondary coils of a transformer is $50$ and $200$ respectively. If the current in the primary coil is $4 A$ , then the current in the secondary coil is

1 A

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2 A

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4 A

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5 A

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Solution

The correct option is A $1 A$
In a transformer:
$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
$\frac{50}{200} = \frac{I_{S}}{4}$
$\Rightarrow I_{S} = 1 A$

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$\frac{N_{P}}{N_{S}} = \frac{I_{S}}{I_{P}}$
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